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# Definite Integrals & Indefinite Integrals Watch

1. Okay, so I'm currently realllllllly confused how to do both.
First, let's go with the Indefinite Integrals:
The question in the book is ∫x√(2x+1) dx, with u = 2x+ 1

I'm getting du/dx as 2, but this states I gotta get x.

Now with definite integrals:
The question is ∫x√(x-1)dx with a limit of 1 to 5

I have no idea where to begin. Pleaase help!
2. Try re arranging U for X and sub that in for X Du/dx is 2 and that part is right.
3. (Original post by Mandy03)
Okay, so I'm currently realllllllly confused how to do both.
First, let's go with the Indefinite Integrals:
The question in the book is ∫x√(2x+1) dx, with u = 2x+ 1

I'm getting du/dx as 2, but this states I gotta get x.

Now with definite integrals:
The question is ∫x√(x-1)dx with a limit of 1 to 5

I have no idea where to begin. Pleaase help!
du/dx = 2 is correct. What do you think dx/du is?

Also, if u = 2x+1, can you rearrange this to get x in terms of u?

The second question is the same idea - you will want to put u = x-1 to simplify the thing inside the square root.

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Updated: June 20, 2016
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